91Ó°ÊÓ

i) Mass of book_{$m=2\text{ k²µ}$}

ii) Height from the ground _{$\left(D\right)=10\text{m}$}

iii) Stretched hand at distance _{$\left(d\right)=1.50\text{m}$}

iv) Gravitational acceleration $\left(D\right)=9.80{\text{m/s}}^{\text{2}}$

The mechanical energy _{$E_{mec}$} of a system is the sum of its kinetic energyand potential energy.

Formula:

i)

ii)

iii)

$h=D−d$

$h=10−1.50$

$h=8.50\text{m}$

From problem 3

_{$U_i=mgD$}

_{$⇒U_i=2×9.80×10$}

_{$⇒U_i=196\text{J}$}

_{$⇒U_f=mgd$}

_{$⇒U_f=2×9.80×1.5$}

_{$⇒U_f=29\text{ J}$}

By using law of conservation of mechanical energy,

_{$K_i+U_i=K_f+U_f$}

_{$⇒0+196=K_f+29$}

_{$\begin{array}{c}{K}_f=196−29\\ =167\text{ J}\end{array}$}

$⇒K_f=\frac{1}{2}m{v}_f^2$

_{$⇒v_f=\sqrt{\frac{2K_f}{m}}$}

_{$⇒v_f=\sqrt{\frac{2×167}{2}}$}

_{$⇒v_f=\sqrt{167}$}

_{$⇒v_f=12.9\text{m/s}$}

If mass is doubled, then$K_f$is also doubled.

$K_f=2×167$

$K_f=m{v}_f^2$

$⇒v_f=\sqrt{\frac{2K_f}{2m}}$

$⇒v_f=\sqrt{\frac{2×2×167}{2×2}}$

$⇒v_f=\sqrt{167}$

$⇒v_f=12.9\text{m/s}$

If$K_i≠0,$we find$K_f=mgh+K_i$ . The value of$K_i$is always positive. This would result in a larger value for$K_f$than in the previous parts, and thus it leads to a larger value for_{$v_f$}.